The photograph in the figure below is a short-time exposure (1/31 s) of a juggle

Question

The photograph in the figure below is a short-time exposure (1/31 s) of a juggler with two tennis balls in the air. (Take the diameter of the tennis ball to be 6.5 cm.)

(a) The tennis ball near the top of its trajectory is less blurred than the lower one. Why is that?
1Because of the very short time exposure the ball on the top appears to be more clear than the ball on the bottom
(b) Estimate the speed of the ball that he is just releasing from his right hand.
2 m/s

(c) Determine how high the ball should have gone above the launch point and compare it to an estimate from the picture. Hint: You have a built-in distance scale if you assume some reasonable value for the height of the juggler.
3 cm

Explanation / Answer

(a) the ball is less blurring beacuse is moves slowly (b) average speed = distance traveled / time taken                                               = 2 ball diametre / time = 2*(6.5*10^-2 m ) /( 1/31 s ) = 4.03 m/s (c) kinematic relation v^2 -u^2 = 2ah v = 0.0m/s u = 4.03 m/s maximum height h = u^2 /2a = (4.03)^2 / (2*9.8 ) = 0.82 m if we assume the height of the juggler is 6 ft (1.8 m) our height h = 0.82 m is good apporximaton as shown in the diagram

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